Indefinite Integration
Integration of trigonometric functions
Grade 12
Question:
<p>The integral \(\int \frac{\sin^2 x\cos^2 x}{(\sin^3 x+\cos^3 x)^2}dx\) equal to</p>
<p>\(\frac{1}{(1+\cot^3 x)}+C\)</p>
<p>\(-\frac{1}{3(1+\tan^3 x)}+C\)</p>
<p>\(\frac{\sin^3 x}{(1+\cos^3 x)}+C\)</p>
<p>\(-\frac{\cos^3 x}{3(1+\sin^3 x)}+C\)</p>
Step-by-Step Solution
Key Concept: Divide numerator and denominator by cos⁶x to convert into a rational function of tan x, then use substitution t = tan x + cot x to simplify the resulting expression.
<p><strong>Step 1:</strong> Divide numerator and denominator by cos⁶x:</p><p>∫ [sin²x cos²x / (sin³x + cos³x)²] dx = ∫ [tan²x sec²x / (tan³x + 1)²] · sec²x dx</p><p>= ∫ [tan²x sec⁴x / (tan³x + 1)²] dx</p><p><strong>Step 2:</strong> Let t = tan x, so dt = sec²x dx. Rewrite as:</p><p>∫ [t² / (t³ + 1)²] dt</p><p><strong>Step 3:</strong> Use the substitution u = t³ + 1, so du = 3t² dt, giving t² dt = du/3:</p><p>∫ [1/(3u²)] du = -1/(3u) + C</p><p><strong>Step 4:</strong> Substitute back u = t³ + 1 = tan³x + 1:</p><p>= -1/[3(tan³x + 1)] + C = -1/[3(sin³x + cos³x)/cos³x] + C</p><p>= -cos³x/[3(sin³x + cos³x)] + C</p><p>∴ Answer: B</p>
Correct Answer: B